Question

Use the Bohr model to find the second longest wavelength of light in the Balmer series for a doubly-ionized Li atom (Z = 3). Recall that the Balmer series corresponds to transitions to the n = 2 level.

Group of answer choices

41.1 nm

117 nm

73.0 nm

54.1 nm

209 nm

Answer 1

Longer the wavelength, smaller should be the energy difference.

for the longest wavelength of Barmer series transition should be from n=3 to n=2

for the second-longest wavelength of Barmer series transition should be from n=4 to n=2

$\frac{1}{\lambda} = R Z^2 (\frac{1}{n_f ^2}- \frac{1}{n_i ^2})$

= (1.097 x 10⁷ m^-1) x 3² x  $(\frac{1}{2^2}- \frac{1}{4 ^2})$

= 1.851 x 10⁷ m^-1

$\lambda=$ 54.1 x 10^-9 m

= 54.1 nm